C4 June 2017 Q2

EdexcelOld spec6 marksBinomial Expansion

2. \[\mathrm{f}(x) = (2 + kx)^{-3}, \quad |kx| < 2, \text{ where } k \text{ is a positive constant}\]

The binomial expansion of \(\mathrm{f}(x)\), in ascending powers of \(x\), up to and including the term in \(x^2\) is \[A + Bx + \frac{243}{16}x^2\] where \(A\) and \(B\) are constants.

(a) Write down the value of \(A\). (1)

(b) Find the value of \(k\). (3)

Mark scheme (a)

Scheme	Marks
\(\left\{(2 + kx)^{-3} = 2^{-3}\left(1 + \dfrac{kx}{2}\right)^{-3} = \dfrac{1}{8}\left[1 + (-3)\left(\dfrac{kx}{2}\right) + \dfrac{(-3)(-3 - 1)}{2!}\left(\dfrac{kx}{2}\right)^2 + \ldots\right]\right\},\ k > 0\)
\(\{A =\}\ \dfrac{1}{8}\) \(\dfrac{1}{8}\) or \(2^{-3}\) or 0.125, clearly identified as \(A\) or as their answer to part (a)	B1 cao
	(1)

Notes

NOTE: IN THIS QUESTION IGNORE LABELLING AND MARK ALL PARTS TOGETHER.

Mark scheme (b)

Scheme	Marks
\(\left(\dfrac{1}{8}\right)\dfrac{(-3)(-4)}{2!}\left(\dfrac{k}{2}\right)^2\) Uses the \(x^2\) term of the binomial expansion to give either* \(\dfrac{(-3)(-4)}{2!}\) or \(\left(\dfrac{k}{2}\right)^2\) or \(\left(\dfrac{kx}{2}\right)^2\) or \(\dfrac{(-3)(-4)}{2}\) or 6* either \((\text{their } A)\dfrac{(-3)(-4)}{2!}\left(\dfrac{k}{2}\right)^2\) or \((\text{their } A)\dfrac{(-3)(-4)}{2!}\left(\dfrac{kx}{2}\right)^2\), where \((\text{their } A) \neq 1\), or \(\dfrac{3}{16}k^2\) or \(\dfrac{3}{16}k^2x^2\) or \((2^{-5})\dfrac{(-3)(-4)}{2!}(kx)^2\) or \((2^{-5})\dfrac{(-3)(-4)}{2!}(k)^2\)	M1 M1 o.e.
\(\left\{\text{So, } \left(\dfrac{1}{8}\right)\dfrac{(-3)(-4)}{2!}\left(\dfrac{k}{2}\right)^2 = \dfrac{243}{16} \Rightarrow \dfrac{3}{16}k^2 = \dfrac{243}{16} \Rightarrow k^2 = 81\right\}\)
So, \(k = 9\) \(k = 9\) cao	A1 cso
Note: \(k = \pm 9\) with no reference to \(k = 9\) only is A0
	(3)

Notes

NOTE: IN THIS QUESTION IGNORE LABELLING AND MARK ALL PARTS TOGETHER.

Note: \((2 + kx)^{-3} = \dfrac{1}{8}\left(1 - \dfrac{3}{2}kx + \dfrac{3}{2}k^2x^2 + \ldots\right) = \dfrac{1}{8} - \dfrac{3}{16}kx + \dfrac{3}{16}k^2x^2 + \ldots\)

Note: Writing down \(\left\{\left(1 + \dfrac{kx}{2}\right)^{-3}\right\} = 1 + (-3)\left(\dfrac{kx}{2}\right) + \dfrac{(-3)(-3 - 1)}{2!}\left(\dfrac{kx}{2}\right)^2 + \ldots\) gets (b) 1^st M1

Note: Writing down \(\left\{(2 + kx)^{-3}\right\} = \dfrac{1}{8}\left(1 + (-3)\left(\dfrac{kx}{2}\right) + \dfrac{(-3)(-3 - 1)}{2!}\left(\dfrac{kx}{2}\right)^2 + \ldots\right)\) gets (b) 1^st M1 2^nd M1 and (c) M1

Note: Writing down \(\left\{(2 + kx)^{-3}\right\} = 2^{-3} + (-3)(2^{-4})(kx) + \dfrac{(-3)(-4)}{2}(2^{-5})(kx)^2\) gets (b) 1^st M1 2^nd M1 and (c) M1

Note: Writing down \(\left\{(2 + kx)^{-3}\right\} = (\text{their } A)\left(1 + (-3)\left(\dfrac{kx}{2}\right) + \dfrac{(-3)(-3 - 1)}{2!}\left(\dfrac{kx}{2}\right)^2 + \ldots\right)\) where \((\text{their } A) \neq 1\), gets (b) 1^st M1 2^nd M1 and (c) M1

Note (b), (c): (their \(A\)) is defined as either

their answer to part (a)
their stated \(A = \ldots\)
their "\(2^{-3}\)" in their stated \(2^{-3}\left(1 + \dfrac{kx}{2}\right)^{-3}\)

Note: Give 2^nd M0 in part (b) if (their \(A\)) = 1

Note: Give M0 in part (c) if (their \(A\)) = 1

Note (b), (c): \({}^{-3}C_0(2)^{-3} + {}^{-3}C_1(2)^{-4}(kx) + {}^{-3}C_2(2)^{-5}(kx)^2\) with the C terms not evaluated gets (b) 1^st M0 2^nd M0 and (c) M0

Mark scheme (c)

Scheme	Marks
"\(\left(\dfrac{1}{8}\right)\)"\((-3)\left(\dfrac{k}{2}\right)\) Uses the \(x\) term of the binomial expansion to give either \((\text{their } A)(-3)\left(\dfrac{k}{2}\right)\) or \((\text{their } A)(-3)\left(\dfrac{kx}{2}\right)\); where \((\text{their } A) \neq 1\), or \((2)^{-4}(-3)(k)\) or \((2)^{-4}(-3)(kx)\) or \(-\dfrac{3k}{16}\)	M1
\(\left\{\text{So, } B = \left(\dfrac{1}{8}\right)(-3)\left(\dfrac{9}{2}\right) \Rightarrow\right\}\ \underline{B = -\dfrac{27}{16}}\) \(-\dfrac{27}{16}\) or \(-1\dfrac{11}{16}\) or \(-1.6875\)	A1 cso
	(2)
	(6 marks)

Notes

Note: Allow M1 for \((\text{their } A)(-3)\left(\dfrac{\text{their } k \text{ from (b)}}{2}\right)\)

Note: Award A0 for \(B = -\dfrac{27}{16}x\)

Note: Allow A1 for \(B = -\dfrac{27}{16}x\) followed by \(B = -\dfrac{27}{16}\) or \(-1\dfrac{11}{16}\) or \(-1.6875\)

Note: \(k = -9\) leading to \(B = \dfrac{27}{16}\) or \(1\dfrac{11}{16}\) or 1.6875 is A0

Note: Give A0 for finding both \(B = -\dfrac{27}{16}\) and \(B = \dfrac{27}{16}\) (without rejecting \(B = \dfrac{27}{16}\)) as their final answer.

Note: The A1 mark in part (c) is for a correct solution only.

Note: Be careful! It is possible to award M0A0 in part (c) for a solution leading to \(B = -\dfrac{27}{16}\). E.g.
\(\mathrm{f}(x) = (2 + kx)^{-3} = 2^{-3}(1 + kx)^{-3} = \dfrac{1}{8}\left(1 + (-3)(kx) + \dfrac{(-3)(-4)}{2!}(kx)^2 + \ldots\right) = \dfrac{1}{8} - \dfrac{3k}{8}x + \dfrac{3k^2}{4}x^2 + \ldots\)
leading to (a) \(A = \dfrac{1}{8}\), (b) \(k = \dfrac{9}{2}\), (c) \(B = -\dfrac{27}{16}\), gets (a) B1, (b) M1M0A0 (c) M0A0